Corollary 4

Let $g\in \mathscr{L}_u$(Sp($f(u)$)). We have $g(f(u)) = (g\circ f)(u)$.

We have $g\circ f\in \mathscr{L}_u$(Sp($u$)) (lemma 5 of IV, p. 184). For all $x$ and $y$ in E, let us denote by $\mu'_{x,y}$ the spectral measure of $(x, y)$ relative to $f(u)$. By the preceding corollary and INT, V, p. 71, § 6, No. 2, th. 1, we have $g\in \mathscr{L}^2$(Sp($f(u)$)$, \mu'_{x,y}$) if and only if $g\circ f\in \mathscr{L}^2$(Sp($u$)$, \mu_{x,y}$), so that the domain of $g(f(u))$ is equal to the domain of $(g\circ f)(u)$. For all $x\in E$ and $y\in$ dom($g(f(u))$), we then have the formula

Sp($f(u)$)

$=\int g f(\mu_{x,y}) =\langle x|(g\circ f)y\rangle$

Sp($f(u)$)

(loc. cit.), whence $g(f(u)) = (g\circ f)(u)$.